congruence modulo twenty four

Printable View

March 22nd 2010, 01:49 AM
EinStone

congruence modulo twenty four

1. if ab ≡ −1 (mod 24), then a + b ≡ 0 (mod 24).
2. If n ≡ −1 (mod 24), then it is not a perfect square, and the number of divisors is even.
3. If n ≡ −1 (mod 24), then 24 divides $\sum_{d|n}d$

I don't need complete proofs, just some hints would be enough.
March 22nd 2010, 02:16 AM
Bacterius

Hello,
(never mind what was there before)
Choose any $a$ . We have $ab \equiv -1 \pmod{24}$ . Therefore, $b \equiv -a^{-1} \pmod{24}$ . Plug this result back into the second condition : $a + b \equiv a - a^{-1} \pmod{24}$ . Now you might want to take a look at what the multiplicative inverses are in $\mathbb{Z}/24\mathbb{Z}$ ;) (it is special)
March 22nd 2010, 02:51 AM
EinStone

Thanks, that proves 1, do you have a hint for 2 as well?
March 22nd 2010, 02:56 AM
Bacterius

Quote:

Originally Posted by EinStone

Thanks, that proves 1, do you have a hint for 2 as well?

Sorry, I haven't had time to look at the other ones and it's bed time for me now :(
Quickly looking, maybe proving that there are no perfect squares of the form $24k - 1$ ? Maybe the prime factorization $24 = 2^3 \times 3$ could help?
March 22nd 2010, 03:45 AM
EinStone

Is this correct for 2.?

Assume n ≡ −1 (mod 24) is a perfect square. Then $n = c^2$ , thus c*c ≡ −1 (mod 24), by 1. we get c + c = 2c ≡ 0 (mod 24).

so $\exists M:$ 2c = M*24 => c= M*12 => $c^2= 144M^2$ , hence $n = c^2 \equiv 0\mod 24$ , a contradiction.

And how to show that there is an even number of divisors?

EDIT: Found 2.
Need proof for 3.

All times are GMT -8. The time now is 09:10 AM.